Subject : S1014 / MECHANICS of MATERIALS
Year : 2008
Mechanical Properties
Session 07-14

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What is Stress ?
Much Work with limited time

High Stress

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What is Stress ?
Less Work with long time

Low Stress

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What is Stress ?
stress is according to strength and failure of solids. The stress field is the
that balance a given set of external tractions and body forces

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Stress stress is according to strength and failure of solids. The stress field is the
that balance a given set of external tractions and body forces

Stress
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Look at the external traction T that represents the force per unit area acting at a given location on the body's surface.

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Stress
Traction T is a bound vector , which means T cannot slide along its line of action or translate to another location and keep the same meaning.
In other words, a traction vector cannot be fully described unless both the force and the surface where the force acts on has been specified.
Given both D F and D s , the traction T can be defined as

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Stress

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Stress
Suppose an arbitrary slice is made across the solid shown in the above figure, leading to the free body diagram shown at right.

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Stress
Surface tractions would appear on the exposed surface, similar in form to the external tractions applied to the body's exterior surface.

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Stress
The stress at point P can be defined using the same equation as was used for T.

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Stress
Stress therefore can be interpreted as internal tractions that act on a defined internal datum plane.
One cannot measure the stress without first specifying the datum plane.

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Stress
Surface tractions, or stresses acting on an internal datum plane, are typically decomposed into three mutually orthogonal components.
One component is normal to the surface and represents
. The other two components are tangential to the surface and represent shear stresses .

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Stress
What is the distinction between normal and tangential tractions, or equivalently, direct and shear stresses?
Direct stresses tend to change the volume of the material (e.g. hydrostatic pressure) and are resisted by the body's bulk modulus (which depends on the Young's modulus and Poisson ratio).

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Stress
What is the distinction between normal and tangential tractions, or equivalently, direct and shear stresses?
tend to deform the material without changing its volume , and are resisted by the body's shear modulus.

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Stress
These nine components can be organized into the matrix:

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Stress where shear stresses across the diagonal are identical ( s xy
= s yx
, s yz s zy
, and s zx
=
= s xz
) as a result of static equilibrium (no net moment).

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Stress
This grouping of the nine stress components is known as the stress tensor (or stress matrix).

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Stress
The subscript notation used for the nine stress components have the following meaning:

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What is Strain?
A propotional
( intensity or degree of distortion )

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What is Strain measure?
a total elongation per unit length of material due to some applied stress.
 

L o

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What are the types of strain ?
1.
2.

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Elastic Strain
Transitory dimensional change that exists only while the initiating stress is applied and dissapears immediately upon removal of the stress .

Elastic Strain
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The applied stresses cause the atom are displaced the same amount and still maintain their relative geometic. When streesses are removed , all the atom return to their original positions and no permanent deformation occurs

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Plastic Deformation a dimentional change that does not dissapear when the initiating stress is removed .
It is usually accompanied by some elastic strain.

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Elastic Strain & Plastic Deformation

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Elastic Strain & Plastic Deformation
Most of Metal material 
At room temperature they have some elasticity , which manifests itself as soon as the slightest stress is applied. Usually, they are also posses some plasticity , but this may not become apparent until the stress has been raised appreciablty.

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Elastic Strain & Plastic Deformation
Most of Metal material 
The magnitude of Plastic strain , when it does appear , is likely to be much greater than that of the elastic strain for a given stress increment

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Constitutive
F
Solid material by force, F, at a point, as shown in the figure .

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Constitutive d
F
Let the deformation at the the point be infinitesimal and be represented by vector d, as shown.
The work done = F .d

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Constitutive z y d
F x
For the general case:
W = F x d x i.e., only the force in the direction of the deformation does work.

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Amount of Work done
Constant Force
If the Force is constant, the work is simply the product of the force and the displacement,
W = Fx
F x
Displacement

Amount of Work done
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Linear Force :
If the force is proportional to the displacement, the work is
1
W

F
F o
2
F o x o x o x
Displacement

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Strain Energy x
F
A simple spring system, subjected to a Force is proportional to displacement x;
F=kx.
Now determine the work done when F= F o
, from before:
W

1
2
F o x o
This energy (work) is stored in the spring and is released when the force is returned to zero

Hooke’s Law
For systems that obey Hooke's law, the extension produced is directly proportional to the load:
F=kx
• where:
– x = the distance that the spring has been stretched or compressed away from the equilibrium position, which is the position where the spring would naturally come to rest (usually in meters),
– F = the restoring force exerted by the material (usually in newtons), and
– K = force constant (or spring constant). The constant has units of force per unit length (usually in newtons per meter).
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Hooke’s Law

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Hooke’s Law

y a
Strain Energy Density a x a
Consider a cube of material acted upon by a force, F x, creating stress s x
=F x
/a 2
W

1
2
F x

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y
F x
Strain Energy Density causing an elastic displacement, d in the x direction, and strain e x
=d/a
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W

1
2
F x
 x
U

1
2 s x a
2 e x a

1
2 s x e x a
3 u

U
V

1
2 s x e x a
3
/ a
3 
1
2 s x e x
Where U is called the Strain Energy , and u is the
Strain Energy Density.

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(a) For a linear elastic material
500 u=1/2(300)(0.0015) N.mm/mm 3
=0.225 N.mm/mm 3
CONTINUED
400
300
200
100
0
0.000
0.002
0.004
0.006
Strain
0.008
0.010

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500
400
300
200
100
(b) Consider elastic-perfectly plastic
CONTINUED u=1/2(350)(0.0018)
+350(0.0022)
=1.085 N.mm/mm 3
0.002
0.004
0.006
Strain
0.008
0.010

Shear Strain Energy y y
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  g xy a
Consider a cube of material acted upon by a shear stress
, t xy causing an elastic shear strain g xy
U

1
2 t xy g y a
3 u

1
2 t xy g xy a
3
/ a
3 
1
2 t xy g xy

Total Strain Energy for a Generalized State of Stress
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
1
2
 s x
 x
 s y
 y
 s z
 z
 t xy g xy
 t yz g
 xx
 yy
 zz g xy g yz g zx






1
 s xx
 
( s yy
E
1
 s
2 ( 1
E
 
)
 
E yy
1
 s zz
E
2 ( 1
 
 
) s s
2 ( 1
E
 
E
) s
( s
( s xy yz zx xx zz
 s
 s
 s zz xx yy


 yz
 t xz g xz


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A
L
Strain Energy for axially loaded bar

F s axial

F
A
;
 
U

1
2
F
 
F
2
L
2 AE
FL
AE
;
F= Axial Force (Newtons, N)
A = Cross-Sectional Area Perpendicular to “F” (mm 2 )
E = Young’s Modulus of Material, MPa
L = Original Length of Bar, mm

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Comparison of Energy Stored in Straight and Stepped bars
L
 a
A F
(a)
U

F
2
L
2 AE

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Comparison of Energy Stored in Straight and Stepped bars

L/2 L/2 b nA A F
(b) U

F
2
L / 2

2 AE
F
2
L / 2
2 nAE

F
2
L
2 AE 2 n n

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What is Torsion ?
an external torque is applied and an internal torque, shear stress, and deformation (twist) develops in response to the externally applied torque.

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What is Torsion ?
For solid and hollow circular shafts, in which assume the material is homogeneous and
isotropic
, that the stress which develop remain within the elastic limits, and that plane sections of the shaft remain plane under the applied torque.

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What is isotropic ?
same

Torsion of shafts
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Shafts are members with length greater than the largest cross sectional dimension used in transmitting torque from one plane to another

Internal Torque
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Consider circular shaft AC subjected to equal and opposite torques T and T’. A cutting plane is passed through the shaft at B.
The FBD for section BC must include the applied torque and elementary shearing forces dF . These forces are perpendicular to radius of the shaft and must balance to maintain equilibrium.
The axis of the shaft is denoted as r
.


M

0 : T
'
   dF

0
T
' 
T : T
   dF
Taking moments about the Axis of the shaft results in dF is related to the shearing stress: dF = t dA
So the applied torque can be related to the shearing stress as
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T
    t dA


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Equation is independent of material model as it represents static equivalency between  shear stress and internal torque on a cross section
T
    t dA


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Shear stress can’t exist on one plane only.
The applied torque produces a shear stress to the axis of the shaft.
The equilibrium require equal stresses on the faces

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What is Torsion ?
To obtain a formula for the relative rotation f
2 internal torque T.
f
1 in terms of the
To obtain a formula for the shear stress t xq torque T.
in terms of the internal f angle of twist

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Shearing Strain

Assume: Material is linearly elastic and isotropic
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T
    t dA

J
  
2 dA Polar moment of inertia for the cross section

Torsion formula
Circular hollow shaft with outer radius R, inner radius r
J


2

R
4
 r
4

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Sign Convention
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Relative rotation f
2
f
1 in terms of the internal torque T.
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Shear stress t x q in terms of the internal torque T.
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Maximum occurs at shaft’s outer radius

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Direction of Shearing

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